Hydrogen peroxide (H₂O₂)decomposes according to the equation H₂O₂(l)$\rarr$ H₂O(l)+ ¹/₂O₂(g). Calculate K_p for this reaction at 25°C.($\Delta$H° = -98.2 kJ/mol, $\Delta$S° = 70.1 J/K·mol)

Question

Quizplus · Accepted Answer

The equilibrium constant can be calculated using the equation:

ΔG° = -RT ln K

where ΔG° is the standard free energy change, R is the gas constant, T is the temperature in Kelvin, and K is the equilibrium constant.

ΔG° = -98.2 kJ/mol = -98,200 J/mol
T = 25°C = 298 K
R = 8.314 J/K·mol

ΔG° = -RT ln K
-98,200 J/mol = -(8.314 J/K·mol)(298 K) ln K
ln K = -46.021
K = e^(-46.021) = 7.5 × 10^-20

Therefore, the answer is D) 7.5 × 10^-20

Hydrogen Peroxide (H2O2)decomposes According to the Equation H2O2(l) →\rarr→ H2O(l)+ 1/2O2(g) Δ\DeltaΔ

Hydrogen Peroxide (H₂O₂)decomposes According to the Equation H₂O₂(l) $\rarr$ H₂O(l)+ ¹/₂O₂(g) $\Delta$